Why is there a large difference in the boiling points of butanal and butan$-1-$ol?
The boiling point of butan$-1-$ol is significantly higher than that of butanal because butan$-1-$ol molecules are associated through intermolecular hydrogen bonding.
In contrast,butanal molecules are held together only by weak dipole-dipole interactions.
This strong intermolecular hydrogen bonding in alcohols requires more energy to break,resulting in a higher boiling point compared to aldehydes of similar molecular mass.
In contrast,butanal molecules are held together only by weak dipole-dipole interactions.
This strong intermolecular hydrogen bonding in alcohols requires more energy to break,resulting in a higher boiling point compared to aldehydes of similar molecular mass.
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Similar Questions
Identify the reagent used in the transformation of a ketone into a semicarbazone.
Match the example given in Column-$I$ with the name of the reaction in Column-$II$.
Column-$I$ (Example) Column-$II$ (Reaction) $(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$ $(1)$ Friedel-Crafts acylation $(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$ $(2)$ $HVZ$ reaction $(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$ $(3)$ Aldol condensation $(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$ $(4)$ Cannizzaro's reaction $(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$ $(5)$ Rosenmund's reduction $(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$ $(6)$ Stephen's reaction
| Column-$I$ (Example) | Column-$II$ (Reaction) |
|---|---|
| $(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$ | $(1)$ Friedel-Crafts acylation |
| $(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$ | $(2)$ $HVZ$ reaction |
| $(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$ | $(3)$ Aldol condensation |
| $(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$ | $(4)$ Cannizzaro's reaction |
| $(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$ | $(5)$ Rosenmund's reduction |
| $(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$ | $(6)$ Stephen's reaction |
Medium
View SolutionGiven below are two statements:
Statement $I$: $2, 6$-diethylcyclohexanone and $6$-methyl-$2$-$n$-propylcyclohexanone are metamers.
Statement $II$: $2, 2, 6, 6$-tetramethylcyclohexanone exhibits keto-enol tautomerism.
Statement $I$: $2, 6$-diethylcyclohexanone and $6$-methyl-$2$-$n$-propylcyclohexanone are metamers.
Statement $II$: $2, 2, 6, 6$-tetramethylcyclohexanone exhibits keto-enol tautomerism.
The starting material is $2,3-\text{dimethylbut-2-ene}$. It undergoes ozonolysis followed by reduction with $Zn/H_2O$ to form product $A$. Product $A$ then reacts with $CH_3MgBr$ followed by hydrolysis to form product $B$. Identify product $B$.
Medium
View SolutionGive the major organic product of the following reaction.
Difficult
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