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Q.(a) Classify the following as primary/secondary/tertiary alcohols: CH3–CH2–CH2–OHCH_3\text{--}CH_2\text{--}CH_2\text{--OH} , and a benzene ring with a -CH2-CH(OH)-CH3 substituent.

(b) Predict the products of the following reactions:
(i) CH3–CH2–CH2–O–CH3+HBr→373 KCH_3\text{--}CH_2\text{--}CH_2\text{--O--}CH_3 + HBr \xrightarrow{373\ K} ?
(ii) phenetole (benzene ring--OC2H5OC_2H_5) →conc. HNO3conc. H2SO4\xrightarrow[\text{conc. } HNO_3]{\text{conc. } H_2SO_4} ?
(c) Give the structure of the products you would expect when Butan-1-ol reacts with the following:
(i) SOCl2SOCl_2
(ii) HCl–ZnCl2HCl\text{--}ZnCl_2 OR
(d) Identify the products A, B, C and D from the following reactions:
(i) chlorobenzene (benzene ring--Cl) →300 atmNaOH, HCl, 623 KA→ΔZn-dustB\xrightarrow[300\ atm]{\text{NaOH, HCl, }623\ K} A \xrightarrow[\Delta]{\text{Zn-dust}} B
(ii) CH3CHO→(ii) H2O/H+(i) CH3MgBrC→443 KH2SO4D+H2OCH_3CHO \xrightarrow[(ii)\ H_2O/H^{+}]{(i)\ CH_3MgBr} C \xrightarrow[443\ K]{H_2SO_4} D + H_2O
(e) Convert cumene to phenol.
(f) Give the structures and IUPAC names of the products expected from the following reactions:
(i) Catalytic reduction of butanal
(ii) Hydration of propene in the presence of dilute sulphuric acid.
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2021Subjective· 5mImportance★★★★★
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This question runs through alcohol/ether/phenol classification, ether cleavage, phenol nitration, and alcohol-to-halide conversions (primary branch), or the industrial phenol-from-cumene route together with a chlorobenzene-to-phenol-to-benzene sequence and a Grignard-then-dehydration sequence (alternative branch).

(a) Classification of alcohols:

  • CH3–CH2–CH2–OHCH_3\text{--}CH_2\text{--}CH_2\text{--OH} (propan-1-ol): the carbon bearing −OH-OH is attached to only one other carbon ⇒\Rightarrow primary (1°) alcohol.
  • Benzene ring—CH2–CH(OH)–CH3CH_2\text{--}CH(OH)\text{--}CH_3 (1-phenylpropan-2-ol): the carbon bearing −OH-OH is attached to two other carbons (the benzylic CH2CH_2 and the terminal CH3CH_3) ⇒\Rightarrow secondary (2°) alcohol.

(b)(i) Ether cleavage: CH3CH2CH2–O–CH3+HBrCH_3CH_2CH_2\text{--O--}CH_3 + HBr (373 K):

On protonation of the ether oxygen, Br−Br^- attacks the less hindered (here, methyl) carbon by an SN2S_N2 pathway, cleaving the C–O bond at the smaller alkyl group:

CH3CH2CH2–O–CH3+HBr→CH3CH2CH2OH+CH3BrCH_3CH_2CH_2\text{--O--}CH_3 + HBr \rightarrow CH_3CH_2CH_2OH + CH_3Br

With excess hot HBr, the resulting propan-1-ol is itself further converted to 1-bromopropane:

CH3CH2CH2OH+HBr→CH3CH2CH2Br+H2OCH_3CH_2CH_2OH + HBr \rightarrow CH_3CH_2CH_2Br + H_2O

(b)(ii) Nitration of phenetole:

The −OC2H5-OC_2H_5 group is strongly activating and ortho/para-directing, so nitration gives a mixture of ortho- and para-nitrophenetole (predominantly para):

C6H5–OC2H5→conc. H2SO4conc. HNO3o-&p-O2N–C6H4–OC2H5+H2OC_6H_5\text{--}OC_2H_5 \xrightarrow[\text{conc. }H_2SO_4]{\text{conc. }HNO_3} o\text{-} \& p\text{-O}_2N\text{--}C_6H_4\text{--}OC_2H_5 + H_2O

(c) Butan-1-ol reactions:

  1. With SOCl2SOCl_2 (best method — clean, only gaseous by-products): CH3CH2CH2CH2OH+SOCl2→CH3CH2CH2CH2Cl+SO2↑+HCl↑CH_3CH_2CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2CH_2Cl + SO_2\uparrow + HCl\uparrow
  2. With Lucas reagent (conc. HClHCl + anhydrous ZnCl2ZnCl_2): primary alcohols react only slowly (requiring heating), via an SN2S_N2-type displacement: CH3CH2CH2CH2OH+HCl→ZnCl2, ΔCH3CH2CH2CH2Cl+H2OCH_3CH_2CH_2CH_2OH + HCl \xrightarrow{ZnCl_2,\ \Delta} CH_3CH_2CH_2CH_2Cl + H_2O Alternative (Or): (d)(i) Chlorobenzene →\rightarrow A →\rightarrow B: Under the Dow's process conditions (fused NaOH, 623 K, 300 atm), chlorobenzene is converted to sodium phenoxide, which on acidification (with HCl) gives phenol: C6H5Cl→then H3O+NaOH, 623 K, 300 atmC6H5OH (A, phenol)C_6H_5Cl \xrightarrow[\text{then }H_3O^+]{NaOH,\ 623\ K,\ 300\ atm} C_6H_5OH\ (A,\ \text{phenol}) Distillation of phenol with zinc dust removes the −OH-OH group (reductive deoxygenation), giving benzene: C6H5OH→ΔZn dustC6H6 (B, benzene)C_6H_5OH \xrightarrow[\Delta]{Zn\ \text{dust}} C_6H_6\ (B,\ \text{benzene}) (d)(ii) CH3CHO→CH_3CHO \rightarrow C →\rightarrow D: Grignard addition of methylmagnesium bromide to acetaldehyde, followed by acidic hydrolysis, gives a secondary alcohol: CH3CHO→(ii) H3O+(i) CH3MgBrCH3–CH(OH)–CH3 (C, propan-2-ol)CH_3CHO \xrightarrow[(ii)\ H_3O^+]{(i)\ CH_3MgBr} CH_3\text{--}CH(OH)\text{--}CH_3\ (C,\ \text{propan-2-ol}) Acid-catalysed dehydration of this secondary alcohol at 443 K gives the alkene: …

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