In the following reaction sequence :
I (\(\mathrm{C_3H_6Cl_2}\)) \(\xrightarrow{\mathrm{KOH(aq)}}\) II \(\xrightarrow[\mathrm{(ii)\ H_2O/H^+}]{\mathrm{(i)\ CH_3MgBr}}\) III \(\xrightarrow{\mathrm{Anhy.\ ZnCl_2 + Con.\ HCl}}\) gives turbidity immediately
The compound I is :
I (\(\mathrm{C_3H_6Cl_2}\)) \(\xrightarrow{\mathrm{KOH(aq)}}\) II \(\xrightarrow[\mathrm{(ii)\ H_2O/H^+}]{\mathrm{(i)\ CH_3MgBr}}\) III \(\xrightarrow{\mathrm{Anhy.\ ZnCl_2 + Con.\ HCl}}\) gives turbidity immediately
The compound I is :
Show Hint
- \(\mathrm{CH_2Cl-CHCl-CH_3}\)
- \(\mathrm{CH_2Cl-CH_2-CH_2Cl}\)
- \(\mathrm{CHCl_2-CH_2-CH_3}\)
- \(\mathrm{CH_3-CCl_2-CH_3}\)
The Correct Option is D
Solution and Explanation
III gives turbidity immediately with anhydrous \(\mathrm{ZnCl_2}\) + conc. HCl (Lucas reagent), so III is a tertiary alcohol.
III is obtained from II by \(\mathrm{CH_3MgBr}\) followed by hydrolysis; a tertiary alcohol arises from a ketone, so II is a ketone. With the C3 skeleton, II = acetone \(\mathrm{CH_3COCH_3}\), and III = \(\mathrm{(CH_3)_3C-OH}\) (tert-butyl alcohol).
Acetone arises from aqueous KOH hydrolysis of a gem-dihalide with both Cl on the middle carbon: \(\mathrm{CH_3-CCl_2-CH_3} \rightarrow [\mathrm{CH_3-C(OH)_2-CH_3}] \rightarrow \mathrm{CH_3COCH_3} + \mathrm{H_2O}\).
Check the others: (3) would give propanal and then 2-butanol (secondary, turbidity only after about 5 minutes); (1) gives propane-1,2-diol (or a hydroxy-carbonyl compound); (2) gives propane-1,3-diol.
So I is 2,2-dichloropropane.
Answer: (4) \(\mathrm{CH_3-CCl_2-CH_3}\)
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