![\[{{[\mathbf{Fe}{{\left( \mathbf{CN} \right)}_{\mathbf{6}}}]}^{\mathbf{4}}}~\mathbf{and}\text{ }{{[\mathbf{Fe}{{({{\mathbf{H}}_{\mathbf{2}}}\mathbf{O})}_{\mathbf{6}}}]}^{\mathbf{2}+}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3b1c1b754a174142f48bf8ad312b8baa_l3.png "Rendered by QuickLaTeX.com")
[Fe( H2O)6 ]2+ and [ Fe( CN)6 ]2+ have H2O and CN- ligands respectively. CN- has a higher CFSE (crystal field splitting energy) than water because it is a strong field ligand. As a result,...
[ Ni (H2O)6 ] is made up of the Ni+2 ion, which has a 3d8 electrical structure. Because H2O is a weak ligand, there are two unpaired electrons in this arrangement that cannot pair up. The d – d...
![\[{{[\mathbf{Cr}{{(\mathbf{N}{{\mathbf{H}}_{\mathbf{3}}})}_{\mathbf{6}}}]}^{\mathbf{3}+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-39b74154417ca7572b9b77e89bc26feb_l3.png "Rendered by QuickLaTeX.com")
In [ Ni ( CN)4 ] 2−, Ni has an oxidation state of +2. Thus, it has d 8configuration.Ni 2+ : Because CN- is a strong field ligand, electrons in 3d orbitals couple. Ni 2+ undergoes dsp2 hybridization...
In the parenthesis are the atomic numbers of the elements are given. Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f7. Yb after forming Yb2+ attains a stable...
The difference in energy between the two levels ( t2g and eg ) that have split from a degenerated d orbital due to the presence of a ligand is known as crystal-field splitting energy. It is...
Ce:1s22s22p63s23p63d104s24p64d105s25p64f15d16s2 Ce3+:1s22s22p63s23p63d104s24p64d105s25p64f1 Magnetic moment can be calculated by the formula: μ = [n(n+2)]1/2 Where, n = number of unpaired...
In the actinoid series, the last element is lawrencium, Lr. The atomic number of the element is 103 and its electronic configuration is [Rn] 5f14 6d1 7s2 . The most common oxidation...
The Spectrochemical series is a set of common ligands arranged in ascending order of their crystal-field splitting energy (CFSE). CFSE readings are higher in strong field ligands. Weak field...
Lanthanoids project 3 different oxidation states which are +2, +3, +4. Among these, the +3 oxidation state is the most common. Lanthanoids exhibit a limited number of oxidation states because of the...
Inner transition metals are the elements in which the last electron is entering in the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block...
An alloy is defined as a solid solution of two or more elements present in a combined form in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys...
(i) Potassium permanganate ( KMnO4 ) can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or...
(i) Potassium dichromate ( K2Cr2O7 ) is prepared from chromite ore (FeCr2O4) in the following steps. Step (1): Preparation of sodium chromate \[4FeC{r_2}{O_4} + 16NaOH + 7{O_2} \to...
(i) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in , the oxidation state of Mn is +7.
![\[\left( i \right)\text{ }{{[Co{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-efe43055f50042489c9d11ee26754efd_l3.png "Rendered by QuickLaTeX.com")
![\[\left( ii \right)\text{ }{{[Co{{F}_{6}}]}^{3-}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a46576bdb43c6d6939fb964ada245115_l3.png "Rendered by QuickLaTeX.com")
(i) Here, the oxidation state of cobalt is +3.Orbitals of Co 3+ ion : Oxalate is a field ligand with a low affinity. As a result, the electrons in the 3d orbital will not pair.Because there are six...
![\[\left( i \right)\text{ }{{[Fe{{\left( CN \right)}_{6}}]}^{4}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c4239d4c20c9394384021fc803fa63f1_l3.png "Rendered by QuickLaTeX.com")
![\[\left( ii \right)\text{ }{{[Fe{{F}_{6}}]}^{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cec302fc55db082efb330ec046bbe8e_l3.png "Rendered by QuickLaTeX.com")
(i) Here, the oxidation state of Fe is +3.Fe 2+ : Electronic configuration is 3d6Orbitals of Fe2+ ion : CN− is a strong field ligand, so it causes the unpaired 3d electrons to pair up: There are six...
![\[{{H}_{2}}S\left( g \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c8bf88a9a8a76dacccc6552af3570fa6_l3.png "Rendered by QuickLaTeX.com")
As a result of the foregoing procedure, the coordination entity obtained is K2[Cu(CN)4 ].The above coordination entity does not ionize to produce Cu2+ ions since it is extremely stable. When...
The presence of [Cu( H2O)4]2+ ions in an aqueous CuSO4 solution gives it a blue colour. (i) As a result, when KF is introduced, H2O ligands are replaced by F- ligands, resulting in green [ CuF4 ]2+...
None of the isomers exhibit optical isomerism.
![\[{{[Co{{(N{{H}_{3}})}_{2}}C{{l}_{2}}\left( en \right)]}^{+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d4533747f86351d7cf03af6d4d2667f_l3.png "Rendered by QuickLaTeX.com")
![\[{{[CoC{{l}_{2}}{{\left( en \right)}_{2}}]}^{+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a90cf5dd5a070919003d67e7a6ddfc4f_l3.png "Rendered by QuickLaTeX.com")
![\[{{[Co(N{{H}_{3}})Cl{{\left( en \right)}_{2}}]}^{2+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4265017b2224d863ef19cf033ef31c23_l3.png "Rendered by QuickLaTeX.com")
(i) (ii)
![\[{{[Cr{{(N{{H}_{3}})}_{2}}C{{l}_{2}}\left( en \right)]}^{+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-826b87fdd17fb106f6e60bd3b147a0cf_l3.png "Rendered by QuickLaTeX.com")
(i) [Cr(C2O4)3]3– (ii) [PtCl2(en)2]2+ Solution: (i) (ii)
( i ) In [ Cr(C2O4)3] 3− no geometric isomers are present because it is a bidentate ligand. ( ii ) In [ Co( NH3)3 Cl3 ]two isomers are possible.
The various types of isomerism that can be observed in coordination compounds are 🙁 i ) Geometrical isomerism : ( ii ) Optical isomerism : (iii) Coordination isomerism :This kind of isomerism...
( i ) Hexaamminenickel(II) chloride ( ii ) Tris(ethane-1, 2-diammine) cobalt(III) ion ( iii ) Tetracarbonylnickel(0)
(i ) Hexaquamanganese(II) ion ( ii) Tetrachloridonickelate(II) ion
(i ) Hexaquatitanium(III) ion ( ii ) Tetraamminichloridonitrito-N-Cobalt(III) chloride
( i ) Hexaamminecobalt(III) chloride ( ii ) Diamminechlorido(methylamine) platinum(II) chloride
( i ) [ Cu (Br)4] 2− ( ii ) [Co ( ONO )( NH3)5] 2+
(i ) K3 [ Cr ( C2O4)3] ( ii ) [ Pt (NH3)6] 4+
(i ) [ Co (NO2) ( NH3)5] 2+ ( ii) [ Co( NH3)6]2 (SO4)3
( i ) [ Pt ( NH3)2Cl2] ( ii) K2[ Ni(CN )4]
( i ) [Zn(OH)4]2 – ( ii ) K2[ Pd Cl4]
Solution: (i) (ii) (iii)
(a) Unidentate ligands: these are ligands with one donor site. Example Cl–, NH3 (b) Bidentate – these are ligands with two donor sites.Example – Ethane-1,2-diamine, Oxalate ion ( C2O42- ) (c)...
(a) Ligands are neutral chemicals or negative ions that are bonded to a metal atom in a coordination entity. Cl-, –OH as an exampleThey are electrically charged radicals or species. ( b )...
When FeSO4is combined with (NH4)2SO4 in a 1: 1 molar ratio, a double salt is formed. FeSO4 (NH4)2SO4 .6H2O. This salt is in charge of supplying Fe 2+ . CuSO 4 in a 1:4 ratio with aqueous ammonia...
( a ) There are two types of valencies in metals: primary and secondary valencies. Primary valencies are satisfied by negative ions, and secondary valencies are filled by both neutral and negative...
(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a...
Cr3+ is the most stable in aqueous solutions owing to a t2g3 configuration.
In the first transition metal series, Cu exhibits +1 oxidation state very frequently because Cu has an electronic configuration of [Ar] 3d 10, that is, the completely filled d-orbital makes it...
It is found that sometimes a relatively less stable oxidation state undergoes an oxidation−reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation....
on platinum surface is zero order reaction. What are the rates of production of 
and 
if 
(i) The ions in d1 configuration tend to lose one more electron to get into stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present...
(i) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This...
![\[\text { For the reaction: } 2 A+B \rightarrow A_{2} B \text { is } k[A][B]^{2} \text { with } k=2.0 \times 10^{-6} \mathrm{~mol}^{-2} L^{2} \mathrm{~s}^{-1} \text {. }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e9fea1165ffad66e28f91d69315a1de4_l3.png "Rendered by QuickLaTeX.com")
![[\mathrm{A}]=0.1 \mathrm{~mol} \mathrm{~L}^{-1},[\mathrm{~B}]=0.2 \mathrm{~mol} \mathrm{~L}^{-1}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d19cf8e34beaaae48c1bfcd427f9aa96_l3.png "Rendered by QuickLaTeX.com")
. Calculate the rate of reaction after [A] is reduced to 
(i) (ii) Solution: (i) (ii)
Solution:
Solution: (i) (ii)
(i) Ethane is generated when methyl bromide is reacted with sodium in the presence of dry ether. The Wurtz reaction is the name for this response. (ii)
(i) Under normal circumstances, chlorobenzene does not hydrolyze. When heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm, it hydrolyzes to produce...
(i) The production of but-1-ene occurs when n-butyl chloride is treated with alcoholic KOH. A dehydrohalogenation process is what this is. (ii) Phenylmagnesium bromide is generated when bromobenzene...
The formula C4H9Br is used to make two main alkyl halides. They're n-butyl bromide and isobutyl bromide, respectively. Compound (a) is therefore either nbutyl bromide or isobutyl bromide. Compound...
KOH almost entirely ionizes in an aqueous solution, yielding OH ions. Because the OH ion is a powerful nucleophile, it causes the alkyl chloride to undergo a substitution reaction, resulting in the...
The symmetrical nature of p-Dichlorobenzene is superior to that of the o- and m-isomers. As a result, it matches the crystal lattice better than the o- and m-isomers. As a result, breaking the...
The production of carbocation occurs during aqueous KOH hydrolysis. If the carbocation is stable, aqueous KOH can easily hydrolyze the compound. Now, C6H5 CH2Cl creates 1o– carbohydrate, whereas...
(i) The nucleophile approaches the carbon atom to which the leaving group is linked in an SN2 reaction. The reactivity for SN2 displacement reduces when the nucleophile is sterically inhibited. The...
Solution: The given reaction is CN- functions as a nucleophile, attacking the carbon atom that the Br is bonded to. The CN– ion is a nucleophile that can attack at both the C and N sites. In this...
Solution:
Solution:
Freon -12 has a variety of applications. Freon-12 (dichlorodifluoromethane, CF2Cl2) is also known as CFC. It's found in deodorants, hair sprays, and other aerosol spray propellants. It's also...
(i) The Cl- atom in chlorobenzene is coupled to an sp2 hybridized carbon atom, whereas it is linked to an sp3 hybridized carbon atom in cyclohexyl chloride. The sp2 hybridized carbon atom is now...
(i) (ii)
(i) (ii)
(i) (ii)
(i) (ii)
(i) (ii)
(i) 1−bromo−1−methylcyclohexane In the given molecule, all β-hydrogens are equivalent. As a result, only one alkene is produced when the given molecule is dehydrogenated. (ii) Different sets of...
(i) Oxidation states : The major oxidation state of lanthanoids is (+3). However, sometimes we also find oxidation states of +2 and +4. This occurs because of the extra stability of fully-filled and...
(i) The order in which the halides react to an alkyl group is constant in the SN2 process. This is because as the size of the halide ion rises, it becomes a better leaving group. R-F << R-Cl...
(i) Electronic configuration: The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2....
From the table shown above we can deduce the following points: Mn shows maximum number of oxidation states, that ranges between +2 to +7.The number of oxidation states increases as we move on from...
The term ambident nucleophile refers to a nucleophile with two nucleophilic sites. These nucleophilic sites are targets for them to attack. For example, the nitrite ion. When the nitrite ion can...
Ions that have electrons in d-orbital will be the only ones that will impart colour and the ions with vacant d-orbital will remain colourless. All other ions, except Sc3+, will be coloured in...
(i) (ii) (iii)
(i) The reduction potentials for the given pairs increase in the given order: Mn2+ / Mn < Cr2+ / Cr < Fe2+ /Fe So, the oxidation of Fe to Fe2+ is not as easy as the...
(i) 1−Bromobutane (ii) 2−Bromobutane (iii) 1−Bromo−2−methylpropane (iv) 2−Bromo−2−methylpropane
C5H10 is a hydrocarbon with the chemical formula CnH2n, which belongs to the CnH2n group of hydrocarbons. As a result, it could be an alkene or a cycloalkane. Hydrocarbon cannot be an alkene because...
Write the ionic equations for the reactions. The preparation of potassium permanganate can be done from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or...
(i) CH3 C (p – Cl C6 H4 )2 CH(Br) CH3 (ii) (CH3 )3 C CH = C Cl C6 H4 I – p (i) 2−Bromo−3, 3−bis(4 − chlorophenyl) butane (ii) 1−chloro−1−(4−iodophenyl)−3,...
In acidic medium, K2Cr2O7 behaves as a very strong oxidising agent. \[{K_2}C{r_2}{O_7} + 4{H_{2}}S{O_4} \to {K_2}S{O_{4}} + C{r_{2}}{(S{O_4})_{3}} + 4{H_2}O + 3[O]\;\] K2Cr2O7 gains electrons...
Potassium dichromate is prepared from iron chromite ore through the following steps: Step 1: Preparation of sodium chromate: \[FeC{r_2}{O_4} + 16NaOH + 7{O_2} \to 8N{a_2}Cr{O_4} + 2F{e_2}{O_3} +...
In transition elements, the oxidation state can vary from +1 to the highest oxidation state by the removal of all their valence electrons. Along with that. in transition elements, the oxidation...
Transition metals are large in size and they also contain many interstitial sites. These interstitial sites can be used to trap atoms of other elements (that have small atomic size), such as H, C, N...
(i) The transition metals generally form coloured compounds.(ii) Transition metals and their many compounds act as good catalyst. (i) Most of the complexes of transition metals are coloured as...
i) Transition metals show paramagnetic behaviour. Paramagnetism is a phenomenon that arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with...
The most common oxidation state in the case of lanthanoids is +3 oxidation state. Thus, Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in...
Transition metals have a partially filled or incomplete d−orbital. Hence, the electronic configuration of transition elements is (n − 1)d1-10 ns0-2. The non-transition elements either do not...
Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled or incomplete d-orbital. These elements are found in the d-block (as the...
Solution: (v) Hemiacetals are a type of α−alkoxyalcohols. A hemiacetal's overall structure. Aldehyde interacts with one molecule of monohydrated alcohol in the presence of dry HCl gas. (vi) Oxime:...
Across a period in the lanthanoid series, there is gradual increase in the atomic number which increases by one. With the increase in atomic number, the number of protons and electrons present in...
The name of the oxometal anions are as follows: (i) Chromate, CrO42- Oxidation state of Cr is + 6. (ii) Permanganate, MnO4- Oxidation state of Mn is + 7. (iii) Vanadate, VO3- Oxidation...
Mn exhibits maximum oxidation states from +2 to +7 and it lies in the first half of the transition series. With an increase in atomic number of the elements, the stability of +2 oxidation state...
In the given table, we can find the oxidation numbers of the elements in the first half of the first row of transition elements with increasing atomic number: Here, we can find that except Sc all...
The electronic configuration of Fe2+ and Mn2+ ions are as follows: Fe2+ is [Ar]18 3d6. Mn2+ is [Ar]18 3d5. As it is known the half and completely filled orbitals have more...
(i) Cl CH2 C ≡ C CH2 Br (iv) ( CCl 3) 3 CCl (i) 1 – Bromo – 4 – chlorobut – 2 – yne (ii) 2−(Trichloromethyl)−1,1,1,2,3,3,3−heptachloropropane
(i) CH3 CH(Cl) CH(Br) CH3 (ii) CH F2 CBr ClF Solution: (i) 2−Bromo−3−chlorobutane (ii) 1−Bromo−1−chloro−1, 2, 2−trifluoroethane
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4 Solution: In CHCl3, the resultant of dipole moments of two C – Cl bonds is countered by the resultant of dipole moments of one CH bond and one C – Cl bond, as seen...
(i) 1 – Bromo – 4 – sec – butyl – 2 – methylbenzene (ii) 1 ,4 – Dibromobut – 2 – ene Solution: (i) (ii)
The electronic configuration of the given elements in their respective oxidation states are: (i) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5 (ii) Th4+:...
The electronic configuration of the given elements in their respective oxidation states are: (i) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar]18 3d7 (ii) Lu2+:...
(i) 2 – Bromobutane (ii) 4 – tert – Butyl -3 –iodoheptane Solution: (i) (ii)
The electronic configuration of the given elements in their respective oxidation states are: (i) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 (ii) Ce+4 :...
(i) 1 -Chloro-4-ethylcyclohexane (ii) 2 – (2 -Chlorophenyl) -1 –iodooctane Solution: (i) (ii)
(i) Cr+3 (ii) Pm+3 The electronic configuration of the given elements in their respective oxidation states are: (i) Cr+3 : 1s2 2s2 2p6 3s2 3p6 3d3 Or, [Ar]18 3d3 (ii)...
(i) 2 -Chloro-3 -methylpentane (ii) p -Bromochlorobenzene Solution: (i) (ii)
(i) m-ClCH2C6H4CH2C(CH3)3 (ii) o-Br-C6H4CH(CH3)CH2CH3 Solution: (i) 1 – Chloromethyl – 3 − ( 2, 2 – dimethylpropyl ) benzene (Primary benzyl halide) (ii) 1 − Bromo − 2 – ( 1 − methylpropyl ) benzene...
(i) CH3C(Cl)(C2H5)CH2CH3 (ii) CH3CH=C(Cl)CH2CH(CH3)2 Solution: (i) 3−Chloro−3–methylpentane (Tertiary alkyl halide) (ii) 3−Chloro−5−methylhex−2−ene (Vinyl halide)
(i) CH3CH(CH3)CH(Br)CH3 (ii) CH3C(C2H5)2CH2Br Solution: (i) 2 − Bromo – 3 – methylbutane (Secondary alkyl halide) (ii) 1 − Bromo − 2 − ethyl − 2 – methylbutane (Primary alkyl...
(i) Cyclopent – 3 – en – 1 – ol (ii) 4-Chloro-3-ethylbutan-1-ol. Solution: (i) (ii)
(i) Cyclohexylmethanol (ii) 3 – Cyclohexylpentan – 3 – ol Solution: (i) (ii)
(i) 1 – Ethoxypropane (ii) 2 – Ethoxy – 3 – methylpentane Solution: (i) structure of 1 – Ethoxypropane : (ii) Structure of 2 – Ethoxy – 3 – methylpentane...
(i) 3 , 5 – Dimethylhexane – 1 , 3 , 5 – triol (ii) 2, 3 – Diethylphenol Solution: (i) (ii)
(i) 2 – Methylbutan – 2 – ol (ii) 1–Phenylpropan–2–ol Solution: (i) (ii)
Solution: The following procedure is involved in the above-mentioned steps of the reaction: Step 1: Protonation Step 2: Forming 2 ° carbonation by eliminating the water molecule Step 3: Rearranging...
(i) (ii) Solution: All the alcohols are formed by the hydration of alkenes in the acidic medium. The addition follows Markownikov’s rule. (i) 1-Methylcyclohexene can be used in the reaction. (ii)...
(i) (ii)
(i) (ii)
Step 1: Protonation of methoxymethane : Step 2: Nucleophilic attack of I – : Step 3: The methanol produced in the previous step is used to combine with another HI molecule, resulting in the...
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in the benzene ring. (i) The presence of...
(i) 1-propoxypropane (ii) methoxybenzene (iii) benzyl ethyl ether
The production of ethers with alcohol dehydration is a bimolecular process ( SN 2) that involves an alcohol molecule attacking a protonated alcohol molecule. The alkyl group should be free in the...
The process of dehydration is used to synthesize 1- propoxypropane from propan –1– ol. In the presence of protic acids (such as H2SO4, H3PO4), propan-1-ol dehydrates to yield 1 – propoxypropane....
Dehydration reaction can be used to synthesise 1- propoxypropane from propan –1– ol. In the presence of protic acids (such as H2SO4, H3PO4), propan-1-ol dehydrates to yield 1 –...
Williamson's synthesis is a flexible approach for making symmetrical and unsymmetrical ethers. However, careful reactant selection is required for the synthesis of unsymmetrical ethers. Because the...
(i) 2-Methoxy-2-methylpropane (ii) 1-Methoxyethane
(i) 1-Propoxypropane (ii) Ethoxybenzene
1 -Ethoxy-4 -4 – dimethyl cyclohexane Ethoxybenzene
(i) 4-Nitroanisole(ii) 1-Methoxypropane
(i)1-Ethoxy-2-methylpropane(ii) 2-Chlorlo-l-methoxy ethane
Because of the presence of the – OH group in ethanol, intermolecular H – bonding occurs, resulting in molecule interaction. To break these hydrogen bonds, more energy is required. Methoxymethane, on...
Because of the presence of the –OH group in ethanol, intermolecular H–bonding occurs, resulting in molecule interaction. To break these hydrogen bonds, more energy is required. Methoxymethane, on...
Solution: (iii) Semicabarbazone: Semicarbazone is made up of aldehydes and ketones, and it is generated when a ketone or aldehyde reacts with semicarbazide. The reaction type is condensation....
Acetal refers to gemdialkoxy alkanes with two groups of alkoxy at the terminal carbon atom. The alkyl group is...
Cyanohydrins are organic compounds having the formula RR'C(OH)CN, where R and R ‘ are alkyl or aryl groups. Aldehydes and ketones react with hydrogen cyanide in the presence of excess sodium cyanide...
(i) Dehydration of propan-2-ol to propene.(ii) Butan-2-one to butan-2-ol. (i) Concentrated H2SO4 or H3PO4 (ii) catalytic hydrogenation or sodium borohydride (NaBH4) or lithium aluminium hydride...
(i) Bromination of phenol to 2,4,6-tribromophenol.(ii) Benzyl alcohol to benzoic acid. (i) Bromine water (ii) Acidified potassium permanganate
(i) Oxidation of primary alcohol to a carboxylic acid.(ii) Oxidation of primary alcohol to aldehyde. (i) NaBH4 or LiAlH4 to acidified KMnO4 (ii) Pyridinium chlorochromate (PCC)
(i) Ethyl magnesium chloride → Propan-1-ol.(ii) Methyl magnesium bromide → 2-Methylpropan-2-ol. (i) When ethyl magnesium chloride reacts with methane, an adduct is formed, resulting in propan–1–ol...
(i) Propene → Propan-2-ol.(ii) Benzyl chloride → Benzyl alcohol. (i) Propene is converted to propan–2–ol when it is treated with water in the presence of an acid catalyst. (ii) If benzyl chloride is...
The mechanism of acid dehydration of ethanol to yield ethene takes place in the following three steps : Step 1 : Formation of ethyl oxonium by protonation of ethanol : Step 2 : Formation of a...
(i) Williamson ether synthesis.(ii) Unsymmetrical ether. (i) Williamson's ether synthesis is the process of obtaining ethers by treating an alkyl halide with a suitable sodium alkoxide. The sodium...
(i) Kolbe's reaction: When sodium phenoxide is heated with C02 at 400°C under 4-7 atmospheres and then acidified, the main result is 2-hydroxybenzoic acid (salicylic acid), with a minor amount of...
(i) Dilute HNO3 with phenol.(ii) Treating phenol with chloroform in the presence of aqueous NaOH. (i) (ii)
(i) (ii)
Because the – OH group works as an electron-donating group, the electron density in the benzene ring increases. The resonance structure of phenol, as seen below, clearly demonstrates this. As a...
An electron-withdrawing group is a nitro – group. The electron density in the O – H bond is reduced by the presence of this group in the ortho position. As a result, it is simpler to give away a...
The acidic nature of phenol can be proven with the two reactions shown below : (i) Reaction with sodium: Phenol reacts with active metals like sodium to liberate Hydrogen gas. (ii) Reaction with...
(i) 1-phenylethanol from a suitable alkene.(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.(iii) pentan-1-ol using a suitable alkyl halide? (i) 1 – phenylethanol can be made by...
Chlorobenzene is mixed with NaOH (at 623 K and 320 atm pressure) to make sodium phenoxide, which is then acidified to produce phenol.
In the mechanism of hydration of ethene to create ethanol, three steps are involved. These are the actions to take: Step 1: Electrophilic action of H3O+ protonates ethene to create carbocation:...
Cumene is first oxidized in the presence of air to produce cumene hydroperoxide, which is then used to make phenol. The cumene hydroxide was then treated with dilute acid to produce phenol and...
o-nitrophenol and p-nitrophenol both have intramolecular H–bonding. Because intermolecular bonding exists in p–nitrophenol, the molecules are tightly bound together. As a result, o – nitrophenol is...
The hydroboration–oxidation reaction is one in which borane is used to initiate the oxidation process. Propan–1–ol, for example, is made by subjecting propene to a hydroboration – oxidation...
Due to the presence of the –OH group, alcohols form H – bonds with water. As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Propanol undergoes intermolecular H-bonding due to the presence of the – OH group. Butane, on the other hand, does not have the same advantages. In order to dissolve the intermolecular hydrogen...
(a) CH3-CH2-CH2-CH2-CH2-OH Pentan – 1 – ol (1 °) (ii) Primary alcohol: Pentan – 1 – ol ; 2 – Methylbutan – 1 – ol ; 3 – Methylbutan – 1 – ol ; 2, 2 – Dimethylpropan – 1 – ol Secondary alcohol:...
(i) 1 – Phenoxyheptane (ii) 2 – Ethoxybutane
(i) 1 – Methoxy – 2 – methyl propane (ii) Ethoxy benzene
(i) 2, 5 – Dimethylphenol (ii) 2, 6 – Dimethylphenol
(i) 2-Methylphenol (ii) 4 – Methyl phenol
(i) Butane – 2, 3 – diol (ii) Propane – 1, 2, 3 – triol
Solutions: (i) 2, 2, 4 -Trimethylpentan – 3 – ol (ii) 5 – Ethylheptane – 2, 4 – diol
(i) tert-Butyl bromide to isobutyl bromide (ii) Aniline to phenylisocyanide Solutions: (i) (ii)
(i) Chloroethane to butane (ii) Benzene to diphenyl Solutions: (i) (ii)
(i) Chlorobenzene to p-nitrophenol (ii) 2-Bromopropane to 1-bromopropane Solutions: (i) (ii)
(i) 2-Chloropropane to 1-propanol (ii) Isopropyl alcohol to iodoform Solution: (i) (ii)
(i) Ethyl chloride to propanoic acid (ii) But-1-ene to n-butyliodide Solution: (i) (ii)
(i) 2-Chlorobutane to 3, 4-dimethylhexane (ii) 2-Methyl-1-propene to 2-chloro-2-methylpropane Solution: (i) (ii)
(i) Ethanol to propanenitrile (ii) Aniline to chlorobenzene Solution: (i) (ii)
(i) Benzene to 4-bromonitrobenzene (ii) Benzyl alcohol to 2-phenylethanoic acid Solution: (i) (ii)
(i) 1-Bromopropane to 2-bromopropane (ii) Toluene to benzyl alcohol Solution: (i) (ii)
(i) Propene to propan-1-ol (ii) Ethanol to but-1-yne Solution: (i) (ii) OR
After 13500C, the standard Gibbs free energy of formation of Al2O3 from Al is lower than that of MgO from Mg. As a result, Al can be expected to be able to reduce MgO at temperatures above 13500C.
Vapour phase refining : It is a process of metal refining in which the metal is converted into a volatile compound, which is then decomposed to obtain the pure metal.(1) To accomplish this, the...
It is the process of employing electricity to refine impure metals. The anode is impure metal, and the cathode is a thin sheet of pure metal in this process. The electrolyte is a metal-specific salt...
The premise behind this approach is that contaminants are more soluble in molten metal than in solid metal. A rotating circular heater slowly glides over an impure metal rod or bar in zone refining....
The anode in aluminium electrometallurgy is graphite, and the cathode iron is graphite lined. Electrolysis liberates O2, which combines with the graphite anode to produce CO2 and CO. If it hadn't...
Chlorine is obtained as a byproduct of the Down process. A fused mixture of CaCl2 and NaCl is electrolyzed at 873 K in this method. At the cathode, sodium is obtained, while Cl2 is released at the...
Ans: The graph of Gibbs energy ∆Gθ versus Temperature for the formation of solid oxides is shown above. This graph shows that if the ∆fGθ of a metal's oxide is more negative than the ∆fGθ of another...
At roughly 1673 K, ZnO is converted to Zn. The Gibbs free energy of formation of CO from C is smaller than the Gibbs free energy of formation of ZnO after 1073 K, and the Gibbs free energy of...
Cr2O3 has a higher value of formation ( −540 kJ mol−1 ) than Al2O3 ( −827 kJ /mol ). Therefore, Al can reduce Cr2O3 to Cr. 2Al + (3/2)O2 ⇒ Al2O3 2Cr + (3/2)O2 ⇒ Cr2O3 Subtracting, 2Al +...
CO will not be able to reduce ZnO to Zn because the standard Gibbs free energy of formation of CO to CO2 is larger than that of Zn to ZnO. As a result, CO-assisted zinc oxide reduction is not...
When working with low-grade copper ores, bacteria or acids are utilised to leach the copper in the presence of air. Copper is introduced to the solution as Cu2+ ions in this method: Cu + 2H+ + ½...
The basic functions of Na3AlF6 (Cryolite) are as follows: 1. Due to the presence of cryolite, the melting point of the solution drops from 2323 K to 1140 K. 2. Cryolite makes alumina a good...
Cu2S and FeS are present in copper matte. When a hot blast of air passes through a molten matte in a silica coated converter, the matte's FeS oxidises to FeO. FeSiO3 (slag) is formed when FeO...
Minerals are substances that contain metals or metal compounds and are found in nature. Ores are rocks and minerals that can be mined for metals in a cost-effective and convenient manner. For...
The iron obtained from a blast furnace is known as pig iron. It is 4 percent carbon and contains minor levels of additional impurities such as Si, P, S, and Mn.Cast iron is created by using a hot...
Calcination is the process of converting carbonate and hydroxide ores to oxides by heating them to temperatures below their melting points while avoiding or limiting the presence of air.This...
The following steps are used to separate alumina from silica in bauxite ore that contains silica:To begin, the powdered ore is digested in a concentrated NaOH solution at 473 523 K and 35 36 bar...
Mond's procedure is a nickel refinement technology. Heat is applied to nickel in the presence of carbon monoxide in this process, resulting in nickel tetracarbonyl, a volatile complex. Ni + 4CO ⇒...
The stationary phase is chosen so that the constituents of the mixture have varying degrees of solubility in it. As a result, distinct elements move at different speeds through the phase, allowing...