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likoan [24]
3 years ago
5

All elements in the same ________________ will have the most similar ___________________ properties.

Chemistry
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

group, chemical

Explanation:

In a periodic table, elements are arranged by electronic configuration, atomic number, and repeating chemical properties.

There are many properties of a periodic table and one of them is that "all the elements present in the same group have most similar chemical properties". This is so because the arrangement of valence electrons in a group is the same which gives the chemical properties to the elements.

For example: Lithium, sodium, potassium, and cesium belong to the same group and have the same chemical properties because they all have 1 valence electron.

Hence, the correct answer is "group, chemical".

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What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

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.

7 0
3 years ago
PLEASE HELP! I DONT UNDERSTAND WHAT TO DO!
goldenfox [79]

Answer:

Na+Cl- + Ag+no3- ---> Na+No3- + Ag+Cl-

A spectator ion is an ion that exists as a reactant and a product in a chemical equation

Explanation:

When a solution of sodium hydroxide, NaOH, is mixed with hydrochloric acid, HCl, the compounds dissociate into the ions Na+, OH-, H+ and Cl-. The hydrogen and hydroxide ions react to form water, but the sodium and chlorine ions stay in solution unchanged.

5 0
3 years ago
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