A substance C4 H10 O yields on oxidation a compound, C 4 H 8 O which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H 2 SO 4 gives C 4 H 8. The structure of the compound is

Question

A substance C4 H10 O yields on oxidation a compound, C 4 H 8 O which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H 2 SO 4 gives C 4 H 8. The structure of the compound is (A) CH 3 CH 2 CH 2 CH 2 OH (B) CH 3 CHOHCH 2 CH 3 (C) ( CH 3)3 COH (D) CH 3 CH 2- O - CH 2 CH 3

Tardigrade Admin · Accepted Answer

C4H8 xleftarrow textConcH2SO2 xleftarrow C4H10O xrightarrow textOxidation C4H8O(R - COCH3) Thus C 4 H 8 O should be CH 3 CH 2 COCH 3, hence C 4 H 10 O should be CH 3 CH 2 CHOH CH 3

Q. A substance $C_{4} H_{10} O$ yields on oxidation a compound, $C_{4} H_{8} O$ which gives an oxime and a positive iodoform test. The original substance on treatment with conc. $H_{2} S O_{4}$ gives $C_{4} H_{8}$ . The structure of the compound is

$C H_{3} C H_{2} C H_{2} C H_{2} O H$

$C H_{3} C H O H C H_{2} C H_{3}$

$(C H_{3})_{3} CO H$

$C H_{3} C H_{2} - O - C H_{2} C H_{3}$

$C_{4} H_{8} Conc H_{2} S O_{2} C_{4} H_{10} O Oxidation C_{4} H_{8} O (R - COC H_{3})$
Thus $C_{4} H_{8} O$ should be $C H_{3} C H_{2} COC H_{3}$ ,
hence $C_{4} H_{10} O$ should be $C H_{3} C H_{2} C H O H C H_{3}$

Q. A substance C4​H10​O yields on oxidation a compound, C4​H8​O which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H2​SO4​ gives C4​H8​. The structure of the compound is

CH3​CH2​CH2​CH2​OH

CH3​CHOHCH2​CH3​

(CH3​)3​COH

CH3​CH2​−O−CH2​CH3​