Hydrocarbons Test

Result

Hydrocarbons Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

In order to prepare 1- chloro-propane which of the following reactants can be employed?

A
Propene and HCl in the presence of peroxide

B
Propene and $$Cl_2$$ followed by treatment with aq.KOH

C
Propanol-1 and $$SOCl_2$$/pyridine

D
Any of the above can be used

Solution

Propanol-1 and $$SOCl_{2}$$ with pyridine as a solvent can be used to prepare 1 - chloro - propane. The by products $$SO_{2} $$ and $$HCl$$ are both gaseous so they go off leaving the actual product.
Question 2
1 / -0

Structure of 'C' is

A

B

C

D

Solution
Question 3
1 / -0

Consider the following sequence of reactions.
$$C_6H_6+CH_3CH=CH_2\overset{H_3PO_4}{\underset{heat}{\rightarrow}}A\overset{1. O_2, heat}{\underset{2. H_3O^+, hear}{\rightarrow}}B+C$$. The products (B) and (C) are.

A
Benzaldehyde and acetaldehyde

B
Benzoic acid and acetic acid

C
Phenol and propionaldehyde

D
Phenol and acetone

Solution

It is a electrophilic addition reaction.The $$O_2$$ ,heat is used for oxidation & it leads to formation of benzaldehyde & acetaldehyde.
Question 4
1 / -0

Choose the appropriate product for this reaction.

A

B

C

D
Question 5
1 / -0

IUPAC name of the compound is:

A
butane

B
$$

\gamma

$$-Butane

C
cyclo-Butane

D
n-Butane

Solution

Since, it is cyclic compound not acyclic straight chain, so 'cyclo' prefix is used for naming and name would be cyclo-butane.
Question 6
1 / -0

The correct IUPAC name of the compound is:

A
5-Ethyl-3,6-dimethyl non-3-ene

B
5-Ethyl-4,7-dimethyl non-3-ene

C
4-methyl-5,7-diethyl oct-2-ene

D
2,4-ethyl-5-methyl oct-2-ene
Question 7
1 / -0

The barrier for rotation about indicated bonds will be maximum in which of these compounds?

A
X

B
Y

C
Z

D
Same in all

Solution
Question 8
1 / -0

The compound $$B$$ formed in the following sequences of reactions is:
$$CH_3CH_2CH_2OH\overset{PCl_3}{\rightarrow}A\overset{Alc. KOH}{\rightarrow}B$$.

A
propyne

B
propene

C
propanol

D
propane

Solution

$$CH_{3}CH_{2}CH_{2}OH\overset{PCl_{3}}{\rightarrow}\underset{propyl\ chloride}{CH_3CH_2CH_2Cl}\overset{alc.KOH}{\rightarrow}\underset{propene}{CH_3CH=CH_2}$$
In the first step the propyl alcohol undergo substitution reaction to form propyl chloride as a proudct which in presence of alc. KOH undergoes elimination reaction to form propene as a product.
Question 9
1 / -0

$$CH_3MgI$$ will give methane with:

A
$$CH_3CH_3$$

B
$$CH_3-CH_2-NH_2$$

C
$$CH_3-CO-CH_3$$

D
all of these

Solution

When the source of protons reacted to Grignard reagent it will definitely give respective alkane, therefore, the only molecule that will react with given Grignard reagent to give methane is ethaneamine.
Question 10
1 / -0

Which of the following reactions will not give propane?

A
$$CH_3CH_2CH_2Cl\overset{Mg/ether}{\underset{H_2O}{\rightarrow}}$$

B
$$CH_3COCl\overset{CH_3MgX}{\underset{H_2O}{\rightarrow}}$$

C
$$CH_3CH=CH_2\overset{B_2H_6}{\underset{CH_3COOH}{\rightarrow}}$$

D
$$CH_3\underset{OH}{\underset{|}{C}}H-CH_3\overset{P/HI}{\rightarrow}$$

Solution

$$(B)$$ will give the above image. So propane cannot be formed.