1. Explain the effects of the following factors on the relative reactivities of Sn1 and Sn2 reaction
a.Nature of the carbon skeleton(primary, secondary, tertiary)
b.Steric hindrance (primary vs hindered primary)
c)Nature of the leaving groups
d)Primary and tertiary vs allylic and benzylic
e)Aliphatic halides vs. alkenyl halide vs. aromatic halide
2. A plot of log k 1/T give s a straight line with the slope equaling to negative Ea/2 3R. Discuss the validity of plotting log t versus 1/T to produce a straight line with a slope equaling to positive Ea/2.3R.
1. a) The nature of the carbon skeleton refers to the number and type of carbon atoms attached to the carbon atom undergoing reaction. In an Sn1 reaction, the reactivity generally increases as the carbon skeleton goes from primary to secondary to tertiary. This is because the stability of the carbocation intermediate formed in the rate-determining step increases with increasing substitution. On the other hand, in an Sn2 reaction, a primary carbon skeleton is more reactive compared to secondary or tertiary carbons. This is because a primary carbon allows for a direct backside attack by the nucleophile without significant steric hindrance.
b) Steric hindrance refers to the crowding of atoms or groups around the carbon atom undergoing reaction. In Sn1 reactions, steric hindrance has a negligible effect because the rate-determining step involves the formation of a carbocation intermediate which is not affected by steric hindrance. However, in Sn2 reactions, steric hindrance hinders the nucleophile from making a backside attack, reducing the rate of the reaction. Therefore, primary carbons with low steric hindrance are more reactive in Sn2 reactions compared to hindered primary carbons.
c) The nature of the leaving group plays a significant role in both Sn1 and Sn2 reactions. A good leaving group should be able to accept an electron pair and leave the carbon atom, preferably with a stable anion formation. In general, the more stable and weaker the leaving group, the faster the reaction. Common leaving groups like halides (e.g., Cl-, Br-, I-) are better than less stable anions (e.g., hydroxide or alkoxide). Better leaving groups increase the reactivity of both Sn1 and Sn2 reactions.
d) Primary carbons are generally more reactive in Sn2 reactions compared to tertiary carbons because primary carbons provide a more accessible site for nucleophilic attack. However, in the case of allylic and benzylic carbons, the reaction rates are increased compared to primary carbons due to the presence of resonance stabilization. Allylic and benzylic carbons have adjacent double bonds or aromatic rings, respectively, which can stabilize the positive charge on the carbon atom. This resonance stabilization enhances the reactivity in both Sn1 and Sn2 reactions.
e) Aliphatic halides refer to halides attached to straight or branched carbon chains. They generally have lower reactivity compared to alkenyl halides (halides attached to carbon-carbon double bonds) and aromatic halides (halides attached to aromatic rings). This is because alkenyl halides and aromatic halides have additional electronic stabilization due to the conjugated nature of the pi system. The presence of pi bonds leads to increased reactivity in both Sn1 and Sn2 reactions.
2. The given plot of log k vs. 1/T with a slope equal to negative Ea/2.3R is consistent with the Arrhenius equation, where k is the rate constant, T is the temperature, Ea is the activation energy, and R is the gas constant. The Arrhenius equation describes the dependence of the rate constant on temperature and activation energy.
However, if we plot log t (where t is the reaction time) versus 1/T, it will not give a straight line with a slope equal to positive Ea/2.3R. This is because the relationship between reaction time and activation energy is not linear. The exponential relationship between rate constant and activation energy in the Arrhenius equation leads to a linear relationship between log k and 1/T, but not between log t and 1/T.
Therefore, plotting log t versus 1/T will not provide a valid approach to determine the activation energy of a reaction. To determine the activation energy, one should plot log k versus 1/T and calculate the slope of the line.