The reaction between 4-methyl-1-pentanol and pbr3 to yield 1-bromo-4-methylpentane is probably:

1 answer:

8 0

It is one of the topic in alcohol reaction and the answer is an SN2-type reaction involving the protonated alkyl dibomophosphite of the alcohol as the substrate. The reaction between 4-methyl-1-pentanol and PBr3to yield 1-bromo-4-methylpentane is probably an SN2-type reaction involving the protonated alkyl dibomophosphite of the alcohol as the substrate.

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The reduction of nitrogen monoxide is described by the following chemical equation: 2H2 (g) +2NO (g) 2H20 ()N2 (g Suppose a two-

Ipatiy [6.2K]

Answer:

Reasonable Second step- $N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)$

Explanation:

The given single step chemical reaction is as follows.

$2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)$

Suppose a two-step mechanism is proposed for this reaction,

The reaction occured in two steps they are as follows.

$Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)$

$Step-2:N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)$

8 0

3 years ago

What is the empirical formula of a carbon-oxygen compound, given that a 95.2 g sample of the compound contains 40.8 g of carbon

algol [13]

Answer:

95.2 - 40.8 = 54.4 g of oxygen

number of moles = mass (g)/ Mr

no. of moles of carbon = 40.8/12 = 3.4

no. of moles of oxygen = 3.4

divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO

Explanation:

hope this helps :)

4 0

3 years ago

Be sure to answer all parts. identify a, b, and c, intermediates in the synthesis of the five-membered ring called an α−methylen

Vanyuwa [196]

The complete reaction along with intermediates is given below, with each step highlighted in different color.

Step 1:
In this step an acidic proton at alpha position is abstracted from lactone moiety and corresponding enolate is formed, which is resonance stabilized. Both structures are shown. In this case LDA (<span>Lithium diisopropylamide) acts as a base.
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Step 2:
The enolate formed attacks on Methyl Iodide, as Iodide being greater in size is a good leaving group and alpha methyl moiety is generated.

Step 3:
This step is acid catalyzed Bromination. Bromine is added at alpha position.

Step 4:
This is elimination reaction and is according to <span>Hofmann's Rule. Here less substituted alkene is generated.</span>