An alkene A C 16 H 16 on ozonolysis gives only one product B C 8 H 8 O - Compound B on reaction with NH 2 OH - H 2 SO 4- - gives N - methyl benzamide the compound -A- is -A-B- - 0 - cH 2- cH - CH - CH 2-0-D-

The correct option is C A B oxime followed by Beckmann's rearrangement So B is ...

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Byju's Answer

Standard XII

Chemistry

Nucleophilic Substitution in Alkyl Halides

An alkene A C...

Question

An alkene (A) $C_{16} H_{16}$ on ozonolysis gives only one product (B) $(C_{8} H_{8} O)$ . Compound (B) on reaction with $\frac{N H_{2} O H}{H_{2} S O_{4}}, Δ$ gives N - methyl benzamide the compound 'A' is -

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Solution

The correct option is B
$A O z o n o l y s i s - ----- \to B N H_{2} O H / H^{+} - ------- \to Δ$ oxime followed by Beckmann's rearrangement

So B is

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An alkene (A) C16H16 on ozonolysis gives only one product (B) (C8H8O). Compound (B) on reaction with NH2OHH2SO4,Δ gives N - methyl benzamide the compound 'A' is -

The correct option is B AOzonolysis−−−−−−→BNH2OH/H+−−−−−−−−→Δ oxime followed by Beckmann's rearrangement So B is

An alkene (A) $C_{16} H_{16}$ on ozonolysis gives only one product (B) $(C_{8} H_{8} O)$ . Compound (B) on reaction with $\frac{N H_{2} O H}{H_{2} S O_{4}}, Δ$ gives N - methyl benzamide the compound 'A' is -

The correct option is B
$A O z o n o l y s i s - ----- \to B N H_{2} O H / H^{+} - ------- \to Δ$ oxime followed by Beckmann's rearrangement

So B is