Question

Reagents such as HCl, HBr, and HOH can add across carbon–carbon double and triple bonds, with H forming a bond to one of the carbon atoms in the multiple bond and Cl, Br, or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions.

Short answer

(a) Product of the addition of water to but-l-en is but--ol.

(b)Product of the addition of hydrogen halide on but-l-ene is-bromobutane.

(c) Product of theaddition hydrogen halide to but-l-yne is2,2-dibromobutane.

(d)Product of the addition of water to1-methylcyclopent-l-ene is1 -methylcyclopentan-l-ol.

(e) Product of the reaction is -chloro--methyl-pentene

Step-by-step solution

Step 1: Addition of water to but-l-en is but-2-ol.

Addition of water or hydrogen halide takes place so that the proton is first added to the end of the double bond that is less substituted.

In this way, a more substituted carbocation is formed, i.e. a more stable carbocation which can then be attacked by water or a halide anion.

Hence The product of the addition of water to but-l-en is but--ol.

Addition of hydrogen halide on but-l-ene is -bromobutane.

Addition of water or hydrogen halide takes place so that the proton is first added to the end of the double bond that is less substituted.

In this way, a more substituted carbocation is formed, i.e. a more stable carbocation which can then be attacked by water or a halide anion.

Hence The product of the addition of hydrogen halide on but-l-ene is -bromobutane.

Step 3: Addition hydrogen halide to but-l-yne is -dibromobutane.

Addition of water or hydrogen halide takes place so that the proton is first added to the end of the triple bond that is less substituted.

In this way, a more substituted carbocation is formed, i.e. a more stable carbocation which can then be attacked by water or a halide anion.

Hence The product of theaddition hydrogen halide to but-l-yne is , 2,2-dibromobutane.

Addition of water to 1-methylcyclopent-l-ene is 1-methylcyclopentan-l-ol.

Addition of water or hydrogen halide takes place so that the proton is first added to the end of the double bond that is less substituted.

In this way, a more substituted carbocation is formed, i.e. a more stable carbocation which can then be attacked by water or a halide anion.

Hence The product of the addition of water to 1-methylcyclopent-l-ene is 1-methylcyclopentan-l-ol.

product of the reaction is 3-chloro-3-methyl-pentene

The following reaction with hydrochloric acid.

-First, the double bond would break because hydrogen (atom from the hydrochloric acid) would take one electron from carbon atom according to the "Markovnikov rule",

-Then the hydrogen atom and the carbon atom would forms bond which would leave the chlorine atom with one extra electron,

-The chlorine atom would give its extra electrons to the other carbon atom and form a single bond there.

Hence the product of the reaction is 3-chloro-3-methyl-pentene