Question

1.   In both the sodium iodide test and the silver nitrate test, why does 2-bromobutane react faster than 2-chlorobutane?
Bromine is a better leaving group since it is a weaker base than chlorine is.

2.   a. Why does benzyl chloride react under both SN1 and SN2 conditions?
            Benzyl chloride is a primary alkyl halide, hence reactive under SN2 conditions.
            The primary carbocation formed due to the departure of Cl- is stabilized by the pi electrons in the benzene ring.
        b. Why is bromobenzene non-reactive under both SN1 and SN2 conditions?   
            For an SN2 reaction to occur, the nucleophile must attack from behind the carbon which is hindered by the bulky ring.
            An SN1 reaction cannot occur since the all the carbons in the benzene ring have sp2 hybridized orbitals.
   

3.      Benzyl chloride and 1-chlorobutane are both primary alkyl chlorides, but in the sodium iodine test, benzyl chloride reacts much faster. Why?
             Benzyl chloride reacts faster because its transition state is more stable than the transition state of 1-chlorobutane.

4.   You wish to substitute the bromine on the following molecules with a nucleophile. Explain whether the given molecule would react by the SN1 or SN2 mechanism and explain why.
        1-methyl-1bromocyclohexane:  SN1 since a tertiary carbocation is relatively stable. SN2 reaction is highly unfavorable at a tertiary carbocation.
        1-bromopropane: SN2 since the nucleophile can readily attack the primary alkyl group. The formation of a primary carbocation intermediate for an SN1 reaction to occur is energetically unfavorable.
        2-bromohexane:  Both SN1 and SN2. The secondary carbocation of the intermediate is relatively stable therefore SN1 reaction is possible. The secondary carbon is also relatively accessible for a back side attack of a nucleophile therefore SN2 reaction is also possible.

5.   The structures below react via the SN1 mechanism. Which would you expect to react slowest and which faster? Explain your prediction.
        5-bromo-1,3-pentadiene: Fastest. Transition state is stabilized by electron delocalization.
        Bromocyclopentane:  Slower.  Transition state  contains a secondary carbocation which is less table than tertiary (c) and electron delocalization stabilized carbocation (a)
        (R)-2-bromo-2-methylhexane:  The transition state is relatively stable due to a tertiary carbocation, therefore (R)-2-bromo-2-methylhexane reacts faster than bromocyclopentane but not as fast as 5-bromo-1,3-pentadiene.
             5-bromo-1,3-pentadiene > (R)-2-bromo-2-methylhexane> Bromocyclopentane

6.   Suppose you find that bromocyclohexane reacts faster than chlorocyclohexane in an SN2 reaction. What reason can you give for this observation?
        Bromine is a better leaving group since it is a weaker base than chlorine is.

7.   1-Chlorobutane (2.5 mL, d=0.886) in 20 mL of acetone is reacted with 90 mL of a 15 wt% solution of NaI in acetone. After workup, you obtain 1.3 g of 1-iodobutane. What is the limiting reagent? What is your % yield?
   g (Acetone)=90mL*d(Acetone)=90mL*0.7925 g/mL=71.325g
   15g(NaI)/(100g (Acetone) )=g(NaI)/(71.325g (Acetone) )?g(NaI)=(15g(NaI)*71.325g(Acetone))/100g(Acetone) = 10.69875g
   g (C_4 H_9 Cl)=V*d=2.5mL*0.886g/mL=2.215g
   2.215g (C_4 H_9 Cl)=0.024mol (C_4 H_9 Cl); 10.7g (NaI)=0.071mol (NaI)
   1-Chlorobutane reacts with one mole of Sodium Iodine to form one mole of 1-iodobutane and one mole of Sodium Chloride. Therefore 0.024mol of 1-Chlorobutane reacts with 0.024mol of Sodium Iodine to form 0.024mol of 1-iodobutane and 0.024mol of sodium chloride. 1-Chlorobutane is the limiting reagent.
   % yield=1.3g/4.4g*100= 29.5%
   
8.      True or False: To promote the SN1 mechanism, you used AgNO3 in a polar, protic solvent. Why is this statement true or false?
        True. Protic solvent is used in SN1 reaction to help in the dissociation of the halogen by stabilization (solvation) of the resulting ions.

9.   True or False: The rate of reaction for the SN2 mechanism is dependent on the concentration of both nucleophile and the electrophile. Justify your answer.
        True. SN2 reaction follows second order kinetics. The reaction occurs when the nucleophile collides (effectively) with the alkyl halide, this collision rate is increased if either one (or both) of the reactants is increased.

Logged

Answer 1

Concepts and reason

The problem is based on the concept of nucleophilic substitution of alkyl halide. There are two mechanisms alkyl halide can proceed nucleophilic substitution namely– ${S_{\rm{N}}}1$ and ${S_{\rm{N}}}2$ .

The ${S_{\rm{N}}}1$ mechanism involves the formation of a carbocation. The halide ion leaves first forming a highly reactive carbocation which is then attacked by a weaker nucleophile.

In an ${S_{\rm{N}}}2$ reaction, a strong nucleophile attacks the carbon attached to halide and pushes it off in one step.

Fundamentals

Alkyl halides which get attacked easily by a nucleophile undergo ${S_{\rm{N}}}2$ reactions, while alkyl halides which can form a stable carbocation undergo ${S_{\rm{N}}}1$ .

The nucleophilicity is affected by various factors like leaving group effects, steric effects and solvent effects.

A good leaving group must be electron withdrawing, stable after leaving and polarizable. Large groups on the carbon center (electrophile) cause hinderance in attack of the nucleophile.

Part 1

In both the tests, 2-chlorobutane reacts slower than 2-bromobutane because chlorine is poor leaving group than bromine.

Part 2.a

Benzyl chloride in ${S_{\rm{N}}}1$ reaction forms a carbocation due to departure of ${\rm{C}}{{\rm{l}}^ - }$ . The primary carbocation formed is stabilized due to the presence of benzene ring. Also, as benzyl chloride is a primary alkyl halide, hence it can undergo ${S_{\rm{N}}}2$ reaction.

Part 2.b

Bromobenzene cannot undergo ${S_{\rm{N}}}1$ reaction as all the carbons in the benzene ring have $s{p^2}$ hybridized orbitals. ${S_{\rm{N}}}2$ reaction cannot occur as the nucleophile cannot approach from behind the carbon due to bulky ring.

Hence, bromobenzene is non-reactive under both ${S_{\rm{N}}}1$ and ${S_{\rm{N}}}2$ conditions.

Part 3

Both benzyl chloride and 1-chlorobutane are primary alkyl chlorides. In the sodium iodide test, benzyl chloride reacts faster due to presence of benzene ring which increases the stability of transition state. Hence, Benzyl chloride reacts much faster than 1-chlorobutane in sodium iodide test.

Ans: Part 1

2-bromobutane react faster than 2-chlorobutane because bromine is better leaving group.

Part 2.a

Benzyl chloride reacts under both ${S_{\rm{N}}}1$ and ${S_{\rm{N}}}2$ conditions.

Part 2.b

Bromobenzene is non-reactive under both S_N1 and S_N2 conditions.

Part 3

Benzyl chloride reacts much faster than 1-chlorobutane in sodium iodide test because its transition state is more stable.