write reactivity order of following:- alkyl halides, Vinyl halides, aryl halide, benzyl halide
Answers
Answer:
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Explanation:
The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing below.
Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
The characteristics noted above lead us to anticipate certain types of reactions that are likely to occur with alkyl halides. The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red.
Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Ionic Nucleophiles
( Weak Bases: I–, Br–, SCN–, N3–,
CH3CO2– , RS–, CN– etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO–, RO– )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2– Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH → CH2=CHCl SN2 substitution. (N ≈ S >>O)
Secondary
R2CH– SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C– E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2– Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Answer:
The correct order of reactivity is :
Benzyl halide > alkyl halide > vinyl halide > acrylic halide
Explanation:
- The benzyl halide compound's halogen is bonded to the carbon that is attached to the benzene ring, and when a carbocation forms at that same carbon, both the carbon and the benzene ring are activated for the attack.
- Alkyl halides are often quite reactive, therefore when the question of which of the vinyl and aryl halides is the most reactive arises, vinyl comes out on top since it is less stable than aryl, and since aryl is the most stable of all these compounds, it is also the least reactive.
- Alkyl halides have a carbon-halogen bond as their functional group, with fluorine, chlorine, bromine, and iodine being the most prevalent halogens.
- These halogens all have electronegativities that are much higher than carbon, with the exception of iodine. As a result, this functional group is polarised, making the halogen nucleophilic and the carbon electrophilic.
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