Question 1

An alkene A on ozonolysis gives a mixture of ethanal and pentan - 3 - one. Write structure and IUPAC name of A.

Accepted Answer

The alkene A is 3-Ethylpent-2-ene. On ozoloysis, it gives a mixture of ethanal and pentan-3-one.

Question 2

Complete the following reactions:A)   CH_3 - CH = CH_2 + HBr \overset { peroxide }{ \rightarrow } ?  B)   H_2 C = CH - CH_3 +HBr { \rightarrow } ?

Accepted Answer

A) Reaction:  CH_3 - CH = CH_2 + HBr \overset { peroxide }{ \rightarrow } CH_2CH_2CH_2Br  B) Reaction:  H_2 C = CH - CH_3 +HBr { \rightarrow } CH_3-CH(Br)-CH_3

Question 3

In the following the most stable conformation of  n- butane is:

Accepted Answer

The anti staggered conformation of n-butane is more stable than gauche staggered and eclipsed conformations of n-butane. In anti staggered n-butane, the methyl groups are placed at a dihedral angle of  180^0 , and the steric hindrance is minimal in anti-form than in gauche form.

Question 4

Arrange the following in decreasing order of their boiling points.(I) n-Butane (II) 2-Methylbutane(III) n-Pentane(IV) 2,2-Dimethylpropane

Accepted Answer

The 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: \underset{III}{n-Pentane} > \underset{II}{2-Methylbutane} > \underset{IV}{2,2-Dimethylpropane} > \underset{I}{n-Butane} Hence,  D  is the correct answer.

Question 5

What is methane? Draw its electron dot structure. Name the  type of bonds formed in this compound. Why are such compounds (i) Are Poor conductors of electricity and(ii) Have low melting and boiling points? What happens when this compound burns in oxygen?

Accepted Answer

Methane is a colourless, odourless and highly flammable gas which is the main component of natural gas.Electron dot structure of methane is- (Image attached)The covalent bonds are present in between four hydrogen atoms and the single carbon atom at the centre of the molecule.(i) Methane is a poor conductor of electricity because in methane all bonds are covalent bonds and therefore no free electrons are present in the molecule that can help in conducting electricity.(ii) Covalent compounds have low intermolecular forces of attraction between the molecules and thus show low melting and boiling points. Since methane is also a covalent compound thus methane has very low melting and low boiling point.When methane is burned in the presence of oxygen, it forms  C{O}_{2}  with water as the product of the reaction. C{H}_{4} + 2{O}_{2} \longrightarrow C{O}_{2} + 2{H}_{2}O

Question 6

CH_{3} CH = CH_{2} + HBr  gives :

Accepted Answer

This is an example of addition reaction. The double bond is broken and hydrogen and bromine are added to the given compound. Bromine will add up on secondary carbon. (Markovnikov's Rule) CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)-CH_3 Option C is correct.

Question 7

Calculate the amount of water (g) product by the combustion of  16 g  of methane.

Accepted Answer

Combustion of methane : C{H}_{4} +2 {O}_{2} \longrightarrow C{O}_{2} + 2 {H}_{2}O Given weight of methane  = 16 g Molecular weight of methane  = 16 g No. of moles of methane in 16g of it  = \cfrac{	ext{given wt.}}{	ext{mol. wt.}} = \cfrac{16}{16} = 1 	ext{ mole} From the reaction,Amount of water produced by the combustion of 1 mole of methane = 2 molesweight of 1 mole of water  = 18 g \Rightarrow  weight of 2 moles of water  = 2 	imes 18 = 36 g Since, the given weight of methane is 16 g, hence the water produced by the combustion of 16 g of methane is 36 g.

Question 8

Write the mechanism of acid dehydration of ethanol to yield ethene.

Accepted Answer

The mechanism of acid dehydration of ethanol to yield ethene is shown above.First step is the protonation of oxygen atom of  -OH  group.Second step is the loss of a molecule of water to form carbonium ion.Last step is deprotonation to form carbon carbon double bond.

Question 9

When benzene reacts with methyl chloride in presence of anhydrous  AlCl_{3} , it gives:

Accepted Answer

when benzene is treated with an alkyl halide in the presence of anhydrous aluminium chloride it gives alkylbenzene. In this reaction hydrogen atom of benzene is replaced by an alkyl group. The reaction is called as Friedel-Craft's alkylation reaction. In this reaction alkyl groups are electrophiles and role of aluminium chloride is to generate these electrophiles. In the given reaction Benzene reacts with methyl chloride to give methylbenzene or toluene.

Question 10

How will you convert ethanoic acid into benzene?

Accepted Answer

I)  CH_{3}COOH+NaOH\rightarrow CH_{3}COONa+H_{2}O (etharoic acid)II)  CH_{3}COONa \xrightarrow[CaO;\Delta ]{NaOH}CH_{4}+Na_{2}CO_{3} III)  CH_{4}+BH_{2}\overset{\Delta }{\rightarrow}CH_{3}Br+HBr IV)  2CH_{3}Br\xrightarrow[cther]{Na}CH_{3}-CH_{3} V)  C_{2}H_{6}+Br_{2}\overset{\Delta }{\rightarrow}CH_{3}-CH_{2}Br VI)  CH_{3}-CH_{2}Br\overset{alc. kOH}{\rightarrow}CH_{2}=CH_{2} VII)  CH_{2}=CH_{2}+Br_{2}\overset{ccl_{4}}{\rightarrow} VIII)  BrCH_{2}-CH_{2}-Br\overset{alc.kOH}{\rightarrow}HC\equiv CH IX)  3CH\equiv CH\xrightarrow[Fe\, tube]{redhst}