230189
The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called
1 aldol condensation
2 Cannizzaro reaction
3 Claisen reaction
4 Reimer-Tiemann reaction
Explanation:
: The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called Reimer-Tiemann reaction.
J and K CET-(2002)
Hydrocarbons
230190
If chloroform is left open in the presence of sunlight, then
1 polymerization takes place
2 explosion takes place
3 poisonous phosgene gas is formed
4 no reaction takes place
Explanation:
: Chloroform in the presence of sunlight and air undergoes reaction with oxygen in air and forms poisonous phosgene gas. \(\underset{\text { Chloroform }}{\mathrm{CHCl}_3}+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\text { Phosgene }}{\mathrm{COCl}_2}+\mathrm{HCl}\)
J and K CET-(2000)
Hydrocarbons
230191
The organic reaction product from the reaction of methyl magnesium bromide and ethyl alcohol is
1 methane
2 ethane
3 propane
4 butane
Explanation:
: Methyl magnesium bromide reacts with ethanol to give methane. Methyl group in the Grignard reagent has a lone pair with a negative charge. This methyl group attacks the hydrogen atom of $-\mathrm{OH}$ group and takes that hydrogen atom towards methyl group. Its form methane. $\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{CH}_2 \mathrm{MaBr}}{\longrightarrow} \underset{\text { Mehlhare }}{\mathrm{CH}_3}-\mathrm{H}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OMgBr}$
J and K CET-(1999)
Hydrocarbons
230192
1-Bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution. The reaction involves elimination of a molecule of
1 $\mathrm{HBr}$
2 $\mathrm{H}_2$
3 $\mathrm{H}_2 \mathrm{O}$
4 $\mathrm{Br}_2$
Explanation:
: 1-bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Alc.KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{HBr}$ In this reaction involves elimination of a molecule of $\mathrm{HBr}$.
230189
The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called
1 aldol condensation
2 Cannizzaro reaction
3 Claisen reaction
4 Reimer-Tiemann reaction
Explanation:
: The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called Reimer-Tiemann reaction.
J and K CET-(2002)
Hydrocarbons
230190
If chloroform is left open in the presence of sunlight, then
1 polymerization takes place
2 explosion takes place
3 poisonous phosgene gas is formed
4 no reaction takes place
Explanation:
: Chloroform in the presence of sunlight and air undergoes reaction with oxygen in air and forms poisonous phosgene gas. \(\underset{\text { Chloroform }}{\mathrm{CHCl}_3}+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\text { Phosgene }}{\mathrm{COCl}_2}+\mathrm{HCl}\)
J and K CET-(2000)
Hydrocarbons
230191
The organic reaction product from the reaction of methyl magnesium bromide and ethyl alcohol is
1 methane
2 ethane
3 propane
4 butane
Explanation:
: Methyl magnesium bromide reacts with ethanol to give methane. Methyl group in the Grignard reagent has a lone pair with a negative charge. This methyl group attacks the hydrogen atom of $-\mathrm{OH}$ group and takes that hydrogen atom towards methyl group. Its form methane. $\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{CH}_2 \mathrm{MaBr}}{\longrightarrow} \underset{\text { Mehlhare }}{\mathrm{CH}_3}-\mathrm{H}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OMgBr}$
J and K CET-(1999)
Hydrocarbons
230192
1-Bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution. The reaction involves elimination of a molecule of
1 $\mathrm{HBr}$
2 $\mathrm{H}_2$
3 $\mathrm{H}_2 \mathrm{O}$
4 $\mathrm{Br}_2$
Explanation:
: 1-bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Alc.KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{HBr}$ In this reaction involves elimination of a molecule of $\mathrm{HBr}$.
230189
The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called
1 aldol condensation
2 Cannizzaro reaction
3 Claisen reaction
4 Reimer-Tiemann reaction
Explanation:
: The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called Reimer-Tiemann reaction.
J and K CET-(2002)
Hydrocarbons
230190
If chloroform is left open in the presence of sunlight, then
1 polymerization takes place
2 explosion takes place
3 poisonous phosgene gas is formed
4 no reaction takes place
Explanation:
: Chloroform in the presence of sunlight and air undergoes reaction with oxygen in air and forms poisonous phosgene gas. \(\underset{\text { Chloroform }}{\mathrm{CHCl}_3}+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\text { Phosgene }}{\mathrm{COCl}_2}+\mathrm{HCl}\)
J and K CET-(2000)
Hydrocarbons
230191
The organic reaction product from the reaction of methyl magnesium bromide and ethyl alcohol is
1 methane
2 ethane
3 propane
4 butane
Explanation:
: Methyl magnesium bromide reacts with ethanol to give methane. Methyl group in the Grignard reagent has a lone pair with a negative charge. This methyl group attacks the hydrogen atom of $-\mathrm{OH}$ group and takes that hydrogen atom towards methyl group. Its form methane. $\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{CH}_2 \mathrm{MaBr}}{\longrightarrow} \underset{\text { Mehlhare }}{\mathrm{CH}_3}-\mathrm{H}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OMgBr}$
J and K CET-(1999)
Hydrocarbons
230192
1-Bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution. The reaction involves elimination of a molecule of
1 $\mathrm{HBr}$
2 $\mathrm{H}_2$
3 $\mathrm{H}_2 \mathrm{O}$
4 $\mathrm{Br}_2$
Explanation:
: 1-bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Alc.KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{HBr}$ In this reaction involves elimination of a molecule of $\mathrm{HBr}$.
230189
The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called
1 aldol condensation
2 Cannizzaro reaction
3 Claisen reaction
4 Reimer-Tiemann reaction
Explanation:
: The reaction of chloroform and sodium hydroxide with phenol to form salicylaldehyde is called Reimer-Tiemann reaction.
J and K CET-(2002)
Hydrocarbons
230190
If chloroform is left open in the presence of sunlight, then
1 polymerization takes place
2 explosion takes place
3 poisonous phosgene gas is formed
4 no reaction takes place
Explanation:
: Chloroform in the presence of sunlight and air undergoes reaction with oxygen in air and forms poisonous phosgene gas. \(\underset{\text { Chloroform }}{\mathrm{CHCl}_3}+\frac{1}{2} \mathrm{O}_2 \rightarrow \underset{\text { Phosgene }}{\mathrm{COCl}_2}+\mathrm{HCl}\)
J and K CET-(2000)
Hydrocarbons
230191
The organic reaction product from the reaction of methyl magnesium bromide and ethyl alcohol is
1 methane
2 ethane
3 propane
4 butane
Explanation:
: Methyl magnesium bromide reacts with ethanol to give methane. Methyl group in the Grignard reagent has a lone pair with a negative charge. This methyl group attacks the hydrogen atom of $-\mathrm{OH}$ group and takes that hydrogen atom towards methyl group. Its form methane. $\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{CH}_2 \mathrm{MaBr}}{\longrightarrow} \underset{\text { Mehlhare }}{\mathrm{CH}_3}-\mathrm{H}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OMgBr}$
J and K CET-(1999)
Hydrocarbons
230192
1-Bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution. The reaction involves elimination of a molecule of
1 $\mathrm{HBr}$
2 $\mathrm{H}_2$
3 $\mathrm{H}_2 \mathrm{O}$
4 $\mathrm{Br}_2$
Explanation:
: 1-bromopropane is converted into propene in the presence of alcoholic $\mathrm{KOH}$ solution $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Alc.KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{HBr}$ In this reaction involves elimination of a molecule of $\mathrm{HBr}$.
