Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.
Answer:Scientific literacy is the knowledge and understanding of scientific concepts and processes required for personal decision making, participation in civic and cultural affairs, and economic productivity. It also includes specific types of abilities.Science literacy is important because it provides a context for addressing societal problems, and because a science- literate populace can better cope with many of its prob- lems and make intelligent and informed decisions that will affect the quality of their lives and those of their children.
Explanation:
Answer:
0.0432M
Explanation:
We begin by writing a balanced equation for the reaction. This is illustrated below:
NaOH + HCl —> NaCl + H2O
From the equation above,
The number of mole of the acid (nA) = 1
The number of mole of the base (nB) = 1
Data obtained from the question include:
Vb (volume of the base) = 54mL
Cb (concentration of the base) = 0.1M
Va (volume of the acid) = 125mL
Ca ( concentration of the acid) =?
Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:
CaVa/CbVb = nA/nB
Ca x 125 / 0.1 x 54 = 1
Cross multiply to express in linear form:
Ca x 125 = 0.1 x 54
Divide both side by 125
Ca = (0.1 x 54) / 125
Ca = 0.0432M
Therefore, the concentration of the acid is 0.0432M
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
Answer:
When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.
Aldolase cleaves a hexose into two trioses.
[See the image attached].
Asterisk indicates the label.
When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.
