Racemic 2-bromobutane

Exercise 8-8 Explain how, in the presence of bromide ion, either enantiomer of 2-bromobutane racemizes (Section 5-1B) in 2-propanone solution at a rate that is first order in Breand first order in 2-bromobutane. 

Reactions between alkyl halides and polar organoalkali compounds often proceed sluggishly. They frequently do not follow SAr2-like mechanistic patterns but rather radical or radical pair pathways.As a corollary, optically active sec-sAky halides such as 2-bromobutane racemize completely when exposed to phenylsodium or -butyllithium. ° To the contrary, predominant inversion is observed with organometallic reagents stabilized either by allylic, propargylic, or benzylic u-delocalization e.g., benzylsodium ) or second-row elements (e.g., l,3-dithian-2-yllithium ). 

In a second example addition of hydrogen bromide converts 2 butene which is achiral to 2 bromobutane which is chiral But as before the product is racemic because... 

Further evidence for a bromine-bridged radical comes from radical substitution of optically active 2-bromobutane. Most of the 2,3-dibromobutane which is formed is racemic, indicating that the stereogenic center is involved in the reaction. A bridged intermediate that can react at either carbon can explain the racemization. When the 3-deuterated reagent is used, it can be shown that the hydrogen (or deuterium) that is abstracted is replaced by bromine with retention of stereochemistry These results are also consistent with a bridged bromine radical. 

It has been shown that a complete shift in stereochemistry of the nucleophilic reactions of (29), with alkyl halides such as 2-bromobutane or cis-2-bromomethoxycyclohexane, from racemization to complete inversion, is induced by increase in the inner-sphere stabilization of the transition state from 0 to 3 kcal mol" This has been ascribed to competition between inner-sphere 5)vr2 and outer-sphere electron-transfer processes the former being extremely sensitive towards inner-sphere stabilization. 

With n dissimilar chiral atoms the number of stereoisomers is 2" and the number of racemic forms is 2" as illustrated below for 2-chloro-3-bromobutan.e (n = 2). The R,S configuration is shown next to... 

Similarly, the addition of HBr to 1-butene produces a racemic mixture of 2-bromobutane. 

Bromination of alkanes follows the same mechanism as chlorination. The only difference is the reactivity of the radical i.e., the chlorine radical is much more reactive than the bromine radical. Thus, the chlorine radical is much less selective than the bromine radical, and it is a useful reaction when there is only one kind of hydrogen in the molecule. If a radical substitution reaction yields a product with a chiral centre, the major product is a racemic mixture. For example, radical chlorination of n-butane produces a 71% racemic mixture of 2-chlorobutane, and bromination of n-butane produces a 98% racemic mixture of 2-bromobutane. 

Stereochemistry of the SnI reactions The SnI reaction is not stereospecific. The carbocation produced is planar and 5p -hybridized. For example, the reaction of (S)-2-bromobutane and ethanol gives a racemic mixture, (S)-2-butanol and (R)-2-butanol. 

If the reagents and reaction conditions are all symmetrical, the product must be a racemic mixture. No optically active material can be created if all starting materials and conditions are optically inactive.70 This statement also holds when one begins with a racemic mixture. Thus racemic 2-butanol, treated with HBr, must give racemic 2-bromobutane. 

One-step addition of H-Br to the top face of the double bond gives (S)-2-bromobutane. Addition of H-Br to the bottom face gives (fi)-2-bromobutane. Since 1-butene is achiral, the probability of addition to either face of the double bond is equal, and the product will be racemic (an equal mixture of enantiomers). 

Problem 8.33 Show reagents and reactions needed to prepare the following compounds from the indicated starting compounds, (a) Acetylene to ethylidene iodide (1,1-diiodoethane). (b) Propyne to isopropyl bromide, (c) 2-Butyne to racemic 2,3-dibromobutane. (d) 2-Bromobutane to trans-2-butene. (e) n-Propyl bromide to 2-hexyne. (/) 1 -Pentene to 2-pentyne. ... 

Most of (Jie biochemical reactions that take place in the body and manj organic reactions in tbe labaratory yield products chirality centers. Fw example, addition of HBr ta l cule. What predictions can we make about the stereochemistry of this rtu-rai product If a single enantianrer is formed, is it A or 5 If a noixtute oS enantiomers id formed, how much -of each In fact, the 2-bromobutane pre duerd is a racemic mixture of R and S enancioiners. Let s sec why. 

To understand why a racemic product results from the reaction of HBr with 1-butene, think about how the reaction occurs. 1-Butene is first proto-nated to yield an intermediate secondary (2") carbocation. Since the trivalent carbon is sp -hybridized and planar, the cation has no chirality centers, has a plane of symmetry, and is achiral. As a result, it can react with Br ion equally well from either the top or the bottom. Attack firom the top leads to (S)-2-bro-mobutane, and attack from the bottom leads to (i )-2-bromobutane. Since both pathways occur with equal probability, a racemic product mixture results (Figure 9.15). 

Now, using the 2-bromobutane enantiomer from exercise 1, make the models of the racemic mixture formed when the bromine is replaced by OH in an SN1 reaction. Visualize the Br leaving first and the water attacking from either side of the carbocation to form the pair of enantiomers. 

As seen in Figure 7.8, the bonds to the positively charged carbon are coplanar and define a plane of symmetry in the carbocation, which is achiral. The rates at which bromide ion attacks the carbocation at its two mirror-image faces are equal, and the product, 2-bromobutane, although chiral, is optically inactive because it is formed as a racemic mixture. 

When a reactant is chiral but optically inactive because it is racemic, any products derived from its reactions with optically inactive reagents will be optically inactive. For example, 2-butanol is chiral and may be converted with hydrogen bromide to 2-bromobutane, which is also chiral. If racemic 2-butanol is used, each enantiomer will react at the same rate with the achiral reagent. Whatever happens to (7 )-( )-2-butanol is mirrored in a corresponding reaction of (5)-(-b)-2-butanol, and a racemic, optically inactive product results. 

A mixture of equal amounts of two enantiomers—such as (/ )-(—)-lactic acid and (6 j-(+)-lactic acid—is called a racemic mixture or a racemate. Racemic mixtures do not rotate the plane of polarized light. They are optically inactive because for every molecule in a racemic mixture that rotates the plane of polarization in one direction, there is a mirror-image molecule that rotates the plane in the opposite direction. As a result, the light emerges from a racemic mixture with its plane of polarization unchanged. The symbol ( ) is used to specify a racemic mixture. Thus, ( )-2-bromobutane indicates a mixture of (-l-)-2-bromobutane and an equal amount of (-)-2-bromobutane. 

---