When 1, 3 - dimethylcyclopentene is heated with ozone and then with zinc and acetic acid, oxidative cleavage leads to keto - aldehyde.
54.
A compound $$\left( X \right)\left( {{C_5}{H_8}} \right)$$ reacts with ammonical $$AgN{O_3}$$ to give a white precipitate, and on oxidation with hot alkaline $$KMn{O_4}$$ gives the acid, $${\left( {C{H_3}} \right)_2}CHCOOH,$$ therefore $$X$$ is –
A
$$C{H_2} = CH - CH = CH - C{H_3}$$
B
$$C{H_3} - CH = CH - C{H_2} - C{H_3}$$
C
$${\left( {C{H_3}} \right)_2}CH - C \equiv CH$$
D
$${\left( {C{H_3}} \right)_2}C = C = C{H_2}$$
Answer :
$${\left( {C{H_3}} \right)_2}CH - C \equiv CH$$
Compound $$X$$ reacts with ammonical $$AgN{O_3}$$ solution, so it must be a terminal alkyne. Formation of $${\left( {C{H_3}} \right)_2}CHCOOH$$ on oxidation of $$X$$ with hot alkaline $$KMn{O_4}$$ further confirms that $$X$$ is $${\left( {C{H_3}} \right)_2}CHC \equiv CH.$$
55.
What is the product when acetylene reacts with the formation of
hypochlorous acid ?
58.
Arrange the following in decreasing order of their boiling points
(i) $$n$$ - Butane
(ii) 2 - Methylbutane
(iii) $$n$$ - Pentane
(iv) 2, 2 - Dimethylpropane
Boiling point of alkanes increases with increase in molecular mass and for the same alkane the boiling point decreases with branching.
Thus, the decreasing order of their boiling points is :
$$\mathop {n{\text{ - Pentane }}}\limits_{\left( {{\text{iii}}} \right)} {\text{ > }}\mathop {{\text{2 - Methylbutane}}}\limits_{\left( {{\text{ii}}} \right)} {\text{ }}$$ $${\text{ > }}\mathop {{\text{2, 2 - Dimethylpropane }}}\limits_{\left( {{\text{iv}}} \right)} $$ $${\text{ > }}\mathop {n{\text{ - Butane }}}\limits_{\left( {\text{i}} \right)} $$
59.
$$HOCl$$ reacts on 3-methyl-2-pentene, the main product will be :