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1 year ago

Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.

1 answer:

7 0

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine ( $Br{2}$ ) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

Initiation reaction: The production of a bromine free radical requires the initiation of heat or light.

$Br - Br$ ⇒ $2Br$ ·

Propagation: This reaction relies heavily on hydrogen. This reaction is impossible if hydrogen is not present. Because tertiary free radicals are more stable than secondary and primary free radicals, they are favoured in this reaction.

Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

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3 years ago

Which organism would have the most variation in the DNA of its offspring?

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2 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

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2 years ago

4. How many moles are in 54.0 grams of Silver?

sineoko [7]

4. Molar mass of silver m Ms~=108 g/mol
Hence there are n=54*(1/108)=0.5 mols of Silver in 54 grams of Silver.

5. 6.3*(108/1)=680.4g

6. Avogadro's number : Na~=6.022×10^23<span>. </span>
6.0*(6.022*10^23/1)=36.132*10^23 atoms

7. Molar mass of Krypton : Mk=84 g/mol
112/84=1.33 moles of Kr

8. 1.93*10^24*(1/(6.022×10^23))=3.2 moles KF

9. Molar mass of Silicon : Ms=28 g/mol
86.2*(1/28)*(6.022×10^23/1)=18.5*10^23 atoms of silicon

10. Molar mass of Magnesium : M1=24 g/mol
4.8*10^24*(1/(6.022×10^23))*(24/1)=191 g Mg

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3 years ago

How would I find the answer?

ELEN [110]

Answer:

Explanation:

The noble gas that is isoelectronic with Br⁻ is krypton.

This is because krypton is the closest noble gas to Br on the periodic table.

Electronic configuration of Bromine is;

2, 8, 18, 7

Br⁻ becomes; 2, 8, 18, 8

Krypton is; 2, 8, 18, 8

4 0

3 years ago