# Can 2‐chloropropane and acetyl chloride give the iodoform test?

Can $$\ce{CH3-CHCl-CH3}$$ and $$\ce{CH3COCl}$$ give the iodoform test?

I thought $$\ce{CH3COCl}$$ having the keto-methyl group would show the iodoform test, while $$\ce{CH3-CHCl-CH3}$$ will not since it's not a secondary alcohol but an alkyl halide.

• acetyl chloride will become acetate under alkaline conditions. Dec 30 '19 at 18:27
• @user55119 the keto methyl group would still be there(?)
– Tapi
Dec 30 '19 at 18:29
• Hydroxide adds to the carbonyl and eliminates chloride. The acetic acid formed is deprotonated by base to acetate. No iodoform reaction of acetate! All carbonyl groups are not created equal. Secondary methyl carbinols are oxidized by I2/NaOH to give the iodoform test via oxidation to acetone. To the extent that SN2 displacement of 2-chloropropane occurs, which is likely very little, 2-propanol is formed. Elimination (E2) is likely the major reaction. Dec 30 '19 at 18:40
• @user55119 Okay, thanks. So, 2-chloropropane also doesn't give the iodoform test.
– Tapi
Dec 30 '19 at 18:46
• 2-Chloropropane itself does not give a positive iodoform test. If you learn the mechanism, then the answer would be self-evident. Dec 30 '19 at 19:03

2-chloropropane will show iodoform test, since through bimolecular substitution reaction $$(\mathrm{S_N2}),$$ chlorine will be replaced by $$\ce{-OH}$$ group. Propan-2-ol so formed will be oxidised to acetone, which will show iodoform test.
Remember that only $$\ce{CH3COR}$$, (where $$\ce{R}$$ can be alkyl or aryl part) or which can be converted to it, give haloform tests, until steric hindrance prevent even some of these compounds to react.